SOA Exam P Sample Question #612. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #612

The median identifies the exponential rate as ln(2)/10. Weighting the two disjoint failure-time windows by their respective payments gives 13.06384325, which rounds to choice A.

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Independent solution

How to solve this Exponential Distribution question

Answer in brief

The median identifies the exponential rate as ln(2)/10. Weighting the two disjoint failure-time windows by their respective payments gives 13.06384325, which rounds to choice A.

Setup

Use the median condition to determine the exponential rate.

\Pr(T>10)=e^{-10c}=0.5

c=\frac{\ln2}{10}

Model

Separate the two payment levels into disjoint time intervals.

E[W]=35\Pr(T\le5)+25\Pr(5<T\le7.5)

F(t)=1-e^{-ct}

Compute

Evaluate the exponential probabilities at both warranty boundaries.

E[W]=35F(5)+25\bigl(F(7.5)-F(5)\bigr)

E[W]=35(1-2^{-1/2})+25(2^{-1/2}-2^{-3/4})

E[W]=13.0638432506\ldots

Answer

The expected warranty payment rounds to 13.06.

\boxed{13.06\quad\text{(A)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

BThis overlaps the two payment windows by treating failures through 7.5 years as eligible for the first payment while also adding the later-window payment.
CThis adds a payment-weighted survival probability rather than using the probability of failure inside each covered interval.
DThis is close to 35P(T>5), which reverses failure and survival at the first warranty boundary and ignores the second payment tier.
EThis extends the second payment beyond the end of the warranty instead of assigning zero after 7.5 years.

Original practice · fully worked

Original variant: capped operating credits

A field instrument remains operational for an exponential time T with mean 4 years. A research program earns 12 credits for each year of operation, but only during the instrument's first three years. The credited amount is proportional for partial years. Calculate the expected total credit.

A
5.668
B
6.332
C
25.326
D
36.000
E
48.000

Variant answer in brief

The credited operating time is min(T,3). Integrating its survival function gives 4(1-e^(-0.75)) years, so the expected credit is 48(1-e^(-0.75))=25.32640547 and choice C is correct.

Setup

Represent the credited duration by the operating lifetime capped at three years.

T\sim\operatorname{Exponential}\!\left(\text{mean }4\right)

C=12\min(T,3)

Model

Use the survival-integral identity for a capped nonnegative duration.

E[\min(T,3)]=\int_0^3\Pr(T>t)\,dt

\Pr(T>t)=e^{-t/4}

Compute

Evaluate the capped lifetime integral and multiply by the annual credit rate.

E[C]=12\int_0^3e^{-t/4}\,dt

E[C]=48(1-e^{-0.75})=25.3264054684\ldots

Answer

The expected total credit rounds to 25.326.

\boxed{25.326\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #612 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.